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## Transportation Problems

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**Transportation Problems**• Transportation is considered as a “special case” of LP • Reasons? • it can be formulated using LP technique so is its solution (to p2)**Here, we attempt to firstly define what are them and then**studying their solution methods: (to p3)**Transportation Problem**• We have seen a sample of transportation problem on slide 29 in lecture 2 • Here, we study its alternative solution method • Consider the following transportation tableau (to p4) (to p6)**Review of Transportation Problem**Warehouse supply of televisions sets: Retail store demand for television sets: 1- Cincinnati 300 A - New York 150 2- Atlanta 200 B - Dallas 250 3- Pittsburgh 200 C - Detroit 200 total 700 total 600 (to p5) LP formulation**A Transportation Example (2 of 3) Model Summary and Computer**Solution with Excel Minimize Z = $16x1A + 18x1B + 11x1C + 14x2A + 12x2B + 13x2C + 13x3A + 15x3B + 17x3C subject to x1A + x1B+ x1 300 x2A+ x2B + x2C 200 x3A+ x3B + x3C 200 x1A + x2A + x3A = 150 x1B + x2B + x3B = 250 x1C + x2C + x3C= 200 xij 0 Exhibit 4.15 (to p3)**Transportation Tableau**• We know how to formulate it using LP technique • Refer to lecture 2 note • Here, we study its solution by firstly attempting to determine its initial tableau • Just like the first simplex tableau! (to p7)**Solution to a transportation problem**• Initial tableau • Optimal solution • Important Notes • Tutorials (to p8) (to p23) (to p43) (to p53)**initial tableau**• Three different ways: • Northwest corner method • The Minimum cell cost method • Vogel’s approximation method (VAM) • Now, are these initial tableaus given us an Optimal solution? (to p9) (to p12) (to p16) (to p7)**Northwest corner method**Steps: • assign largest possible allocation to the cell in the upper left-hand corner of the tableau • Repeat step 1 until all allocations have been assigned • Stop. Initial tableau is obtained Example (to p10)**Northeast corner**Step 1 Step 2 Max (150,200) 150 0 0 -- -- 150 -- -- 50 125 -- (to p11) 0 150**Initial tableau of NW corner method**• Repeat the above steps, we have the following tableau. • Stop. Since all allocated have been assigned (to p8) Ensure that all columns and rows added up to its respective totals.**The Minimum cell cost method**Here, we use the following steps: Steps: Step 1 Find the cell that has the least cost Step 2: Assign as much as allocation to this cell Step 3: Block those cells that cannot be allocated Step 4: Repeat above steps until all allocation have been assigned. Example: (to p13)**Step 1 Find the cell that has the least costStep 2: Assign**as much as allocation to this cell Step 1: Step 2: 200 The min cost, so allocate as much resource as possible here Step 3 (to p14)**Step 3: Block those cells that cannot be allocatedStep 4:**Repeat above steps until all allocation have been assigned. Second iteration, step 4 Step 3: -- -- (to p15) 75 200 0**The initial solution**• Stop. The above tableau is an initial tableau because all allocations have been assigned (to p8)**Vogel’s approximation method**Operational steps: Step 1: for each column and row, determine its penalty cost by subtracting their two of their least cost Step 2: select row/column that has the highest penalty cost in step 1 Step 3: assign as much as allocation to the selected row/column that has the least cost Step 4: Block those cells that cannot be further allocated Step 5: Repeat above steps until all allocations have been assigned Example (to p17)**subtracting their two of their least**cost Step 1 (8-6) (11-7) (5-4) (to p18) (6-4) (8-5) (11-10)**Steps 2 & 3**Step 2: Highest penalty cost (to p19) Step 3: this has the least cost**Step 4**(to p20) --- ---**Step 5Second Iteration**--- (to p21) --- ---**3rd Iteration of VAM**--- --- (to p22) --- ---**Initial tableau for VAM**(to p8)**Optimal solution?**Initial solution from: Northeast cost, total cost =$5,925 The min cost, total cost =$4,550 VAM, total cost = $5,125 (note: here, we are not saying the second one always better!) It shows that the second one has the min cost, but is it the optimal solution? (to p24)**Solution methods**• We need a method, like the simplex method, to check and obtain the optimal solution • Two methods: • Stepping-stone method • Modified distributed method (MODI) (to p25) (to p30) (to p7)**Stepping-stone method**Let consider the following initial tableau from the Min Cost algorithm There are Non-basic variables These are basic variables (to p26) Question: How can we introduce a non-basic variable into basic variable?**Introducea non-basic variable into basic variables**• Here, we can select any non-basic variable as an entry and then using the “+ and –” steps to form a closed loop as follows: Then we have let consider this non basic variable (to p27)**Stepping stone**- + + - (to p28) The above saying that, we add min value of all –ve cells into cell that has “+” sign, and subtracts the same value to the “-ve” cells Thus, max –ve is min (200,25) = 25, and we add 25 to cell A1 and A3, and subtract it from B1 and A3**Stepping stone**The above tableaus give min cost = 25*6 + 120*10 + 175*11 175*4 + 100* 5 = $4525 We can repeat this process to all possible non-basic cells in that above tableau until one has the min cost! NOT a Good solution method (to p29)**Getting optimal solution**• In such, we introducing the next algorithm called Modified Distribution (MODI) (to p24)**Modified distributed method (MODI)**• It is a modified version of stepping stone method • MODI has two important elements: • It determines if a tableau is the optimal one • It tells you which non-basic variable should be firstly considered as an entry variable • It makes use of stepping-stone to get its answer of next iteration • How it works? (to p31)**Procedure (MODI)**Step 0: let ui, v , cij variables represent rows, columns, and cost in the transportation tableau, respectively Step 1: (a) form a set of equations that uses to represent all basic variables ui + vj = cij (b) solve variables by assign one variable = 0 Step2: (a) form a set of equations use to represent non-basic variable (or empty cell) as such cij – ui – vj = kij (b) solve variables by using step 1b information Step 3: Select the cell that has the most –ve value in 2b Step 4: Use stepping-stone method to allocate resource to cell in step 3 Step 5: Repeat the above steps until all cells in 2a has no negative value (to p24) Example (to p32)**MODI**Consider to this initial tableau: Step 0: let ui, v , cij variables represent rows, columns, and cost in the transportation tableau, respectively (to p33)**Step 0**Step 1: (a) form a set of equations that uses to represent all basic variables ui + vj = cij (to p34) C3A**ui + vj = cij**(to p35) (b) solve variables by assign one variable = 0**Set one variable = 0**Because we added an non-basic variable (to p36) Step2: (a & b)**Step2: (a & b)**Note this may look difficult and complicated, however, we can add these V=values into the above tableau as well (to p37)**Step2: (a & b), alternative**-1 -1 +2 +5 (to p38) 6-0-7 Step 3: Select the cell that has the most –ve value in 2b**Step3**-1 -1 +2 +5 (to p39) Select either one, (Why?) These cells mean, introduce it will reduce the min z to -1 cost unit**Step 4: Use stepping-stone method**- + + - From here we have …. (to p40)**Step 4: Use stepping-stone method**(to p41) Step 5: we repeat steps 1-4 again for the above tableau, we have**Step 5**(to p42)**Step 5 cont**All positives STOP (to p31)**Important Notes**• When start solving a transportation problem using algorithm, we need to ensure the following: • Alternative solution • Total demand ≠ total supply • Degeneracy • others (to p44) (to p45) (to p48) (to p52) (to p7)**Alternative solution**• When on the following k = 0 cij – ui – vj = kij Why? (to p43)**Total demand ≠ total supply**Note that, total demand=650, and total supply = 600 How to solve it? We need to add a dummy row and assign o cost to each cell as such .. (to p46)**Dd≠ss**Extra row, since Demand > supply Other example …… (to p47)**Dd≠ss**Extra column is added (to p43)**Degeneracy**(to p49) Example ….. ie, basic variables in the tableau**Degeneracy**• m rows + n column – 1 = the number of cells with allocations • + 3 - 1 = 5 • It satisfied. • If failed? …… considering ….. (five basic variables, and above has 5 as well!) (to p50)**Degeneracy**(note above has only 4 basic variable only!) If not matched, then we select an non-basic variable with least cheapest cost and considered it as a new basic variable with assigned 0 allocation to it (to p51)